Giải thích các bước giải:

a.Xét $\Delta ABH,\Delta ACH$ có:

Chung $AH$
$HB=HC$ vì $H$ là trung điểm $BC$
$AB=AC$

$\to\Delta ABH=\Delta ACH(c.c.c)$

b.Xét $\Delta AHB,\Delta MHC$ có:

$HB=HC$

$\widehat{AHB}=\widehat{MHC}$

$HA=HM$

$\to\Delta AHB=\Delta MHC(c.g.c)$

$\to \widehat{ABH}=\widehat{HCM}\to AB//CM$

3.Xét $\Delta BKC,\Delta BKD$ có:

Chung $BK$

$\widehat{BKC}=\widehat{BKD}=90^o$ vì $BK\perp AC$

$KC=KD$
$\to\Delta BKC=\Delta BKD(c.g.c)$

$\to\widehat{CBK}=\widehat{DBK}$

$\to BK$ là phân giác $\widehat{CBD}$

4.Từ câu a

$\to\widehat{ABH}=\widehat{ACH}$

Mặt khác theo câu c $\to\widehat{BDK}=\widehat{BCK}, BC=BD$

$\to\widehat{ABH}=\widehat{BDK}$

$\to 180^o-\widehat{ABH}=180^o-\widehat{BDK}$

$\to \widehat{EBC}=\widehat{ADB}$

Xét $\Delta ABD,\Delta ECB$ có:
$BC=BD$

$\widehat{ADB}=\widehat{CBE}$

$AD=BE$

$\to\Delta ABD=\Delta ECB(c.g.c)$

$\to AB=CE$

Mà $AB=AC\to CE=CA$